What is the integral of 4e^x(e^x-1) ?

The integral of 4e^x(e^x-1) is 4(1/2 e^{2x}-e^x)+C

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integral of 4e^x(e^x-1)

\int 4 e^{x} (e^{x} - 1) d x

4 (\frac{1}{2} e^{2 x} - e^{x}) + C

Solution steps

\int 4 e^{x} (e^{x} - 1) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int e^{x} (e^{x} - 1) d x

Apply u-substitution

: \int u - 1 d u

= 4 \cdot \int u - 1 d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 4 (\int u d u - \int 1 d u)

\int u d u = \frac{u ^{2}}{2}

\int 1 d u = u

= 4 (\frac{u ^{2}}{2} - u)

Substitute back

u = e^{x}

= 4 (\frac{( e ^{x} ) ^{2}}{2} - e^{x})

Simplify

4 (\frac{( e ^{x} ) ^{2}}{2} - e^{x}) : 4 (\frac{1}{2} e^{2 x} - e^{x})

= 4 (\frac{1}{2} e^{2 x} - e^{x})

Add a constant to the solution

= 4 (\frac{1}{2} e^{2 x} - e^{x}) + C

\frac{2 x ^{9} - 5 x + x}{x ^{4}}

x \to 1 + lim (\frac{1}{x - 1})

(e^{2 x} + 4) (\frac{d y}{d x}) = y

\frac{d}{d x} (2 x - ln (x))

y^{^{'}} = - k y

What is the integral of 4e^x(e^x-1) ?
The integral of 4e^x(e^x-1) is 4(1/2 e^{2x}-e^x)+C